Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<stdio.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int bfs1[100000];
int bfs2[100000];
int v1[100000];
int v2[100000];
int A[100000];
int B[100000];
vector<int> g[100000];
int main(){
	int a,b,c,d;
	scanf("%d%d%d%d",&a,&b,&c,&d);
	c--;d--;
	for(int i=0;i<b;i++){
		int e,f;
		scanf("%d%d",&e,&f);
		e--;f--;
		g[e].push_back(f);
		g[f].push_back(e);
	}
	queue<int> Q;
	Q.push(c);
	for(int i=0;i<a;i++)bfs1[i]=bfs2[i]=-1;
	bfs1[c]=0;
	while(Q.size()){
		int at=Q.front();
		Q.pop();
		for(int i=0;i<g[at].size();i++){
			if(!~bfs1[g[at][i]]){
				bfs1[g[at][i]]=bfs1[at]+1;
				Q.push(g[at][i]);
			}
		}
	}
	queue<int> P;
	P.push(d);
	bfs2[d]=0;
	while(P.size()){
		int at=P.front();
		P.pop();
		for(int i=0;i<g[at].size();i++){
			if(!~bfs2[g[at][i]]){
				bfs2[g[at][i]]=bfs2[at]+1;
				P.push(g[at][i]);
			}
		}
	}
	for(int i=0;i<a;i++){
		if(~bfs1[i]){
			A[bfs1[i]]++;
		}
		if(~bfs2[i]){
			B[bfs2[i]]++;
		}
	}
	int dis=bfs1[d];
	long long ret=0LL;
	for(int i=0;i<dis-1;i++){
//		printf("%d %d\n",A[i],B[i]);
		ret+=(long long)A[i]*B[dis-2-i];
	}
	printf("%lld\n",ret);
}